🧠:<think>Okay, so I've got this problem about an aerospace engineering student designing a new football. It's split into two parts. Let me try to tackle each part step by step.Starting with part 1: The football is a prolate spheroid with semi-major axis 14 cm and semi-minor axis 7 cm. I need to derive the equation governing its rotational dynamics, considering moments of inertia I₁ and I₃, and the aerodynamic torque τ.Hmm, rotational dynamics... I remember that for a rigid body, the rotational motion is governed by Euler's equations. Since the football is rotating about its principal axis, maybe the equations simplify. Let me recall Euler's equations:For a rigid body with principal moments of inertia I₁, I₂, I₃, the equations are:I₁ * (dω₁/dt) = (I₂ - I₃)ω₂ω₃ + τ₁I₂ * (dω₂/dt) = (I₃ - I₁)ω₃ω₁ + τ₂I₃ * (dω₃/dt) = (I₁ - I₂)ω₁ω₂ + τ₃But in this case, the football is rotating about its principal axis, so I think only one component of angular velocity is non-zero. If it's rotating about the principal axis, say the semi-major axis, which is the longest axis, then ω₁ and ω₂ would be zero, and only ω₃ is non-zero. So, let's see.Wait, the football is a prolate spheroid, which is longer along the semi-major axis. So, the principal axes would be along the semi-major (I₃) and semi-minor (I₁) axes. So, if it's rotating about its principal axis, which is the semi-major axis, then ω₁ and ω₂ are zero, and only ω₃ is non-zero. Therefore, the Euler equations simplify.Looking at Euler's equations, if ω₁ = ω₂ = 0, then all the cross terms (like ω₂ω₃) would be zero. So, the equations reduce to:I₁ * (dω₁/dt) = τ₁I₂ * (dω₂/dt) = τ₂I₃ * (dω₃/dt) = τ₃But since the football is symmetric about the semi-major axis, I think I₁ = I₂. Because for a prolate spheroid, the moments of inertia about the semi-minor axes are equal. So, I₁ = I₂, and I₃ is different.Therefore, if I₁ = I₂, then the first two Euler equations become:I₁ * (dω₁/dt) = τ₁I₁ * (dω₂/dt) = τ₂And the third equation is:I₃ * (dω₃/dt) = τ₃But since the football is rotating about its principal axis, I think there might not be any external torque in the direction of ω₃, unless there's aerodynamic torque. Wait, the problem mentions aerodynamic torque τ. So, I need to consider that.Assuming that the aerodynamic torque is acting in the direction perpendicular to the rotation axis? Or maybe along the rotation axis? Hmm, I'm not entirely sure. But in the case of a spinning football, the Magnus effect comes into play, which causes a lateral force. But in terms of torque, if the football is spinning, the aerodynamic torque might be trying to change the angular velocity.Wait, but the football is a prolate spheroid, so it's more streamlined along the semi-major axis. If it's rotating about its principal axis, which is the semi-major axis, then perhaps the aerodynamic torque is trying to maintain or change that spin.Alternatively, maybe the torque is due to the pressure distribution around the football as it moves through the air, causing a torque that affects the angular velocity.But I'm not entirely certain about the direction of the torque. Maybe I should look up the typical aerodynamic torque on a prolate spheroid. But since I can't look things up right now, I need to think.In the case of a spinning sphere, the Magnus effect causes a force perpendicular to the direction of motion and the axis of rotation. But for a prolate spheroid, the aerodynamics might be different because of its shape.Alternatively, maybe the torque is due to the difference in pressure on the front and back of the football as it moves through the air. If it's spinning, the pressure distribution could cause a torque.But perhaps for simplicity, the problem is assuming that the aerodynamic torque is acting along the principal axis, trying to change the angular velocity. So, in that case, the torque τ is along the ω₃ direction.Therefore, the equation governing the rotational dynamics would be:I₃ * (dω₃/dt) = τBut wait, is that all? Or is there more to it?Wait, if the football is moving through the air, there might be other torques acting on it. But since it's a rigid body, and assuming that the only torque is the aerodynamic torque, then yes, the equation would be I₃ * (dω₃/dt) = τ.But I think I need to express τ in terms of other variables. The problem mentions aerodynamic torque, so maybe it's related to the angular velocity and the drag force.Alternatively, perhaps the torque is proportional to the angular velocity, similar to how damping works. So, τ = -k * ω, where k is some constant related to the aerodynamics.But the problem doesn't specify the form of the torque, just mentions to account for it. So, maybe I can leave it as τ in the equation.Therefore, the equation governing the rotational dynamics is:I₃ * (dω/dt) = τBut wait, I should check the moments of inertia for a prolate spheroid. The moment of inertia about the semi-major axis (I₃) is different from the moment about the semi-minor axes (I₁ and I₂). For a prolate spheroid, the moments of inertia are given by:I₁ = I₂ = (2/5) m (a² + b²)I₃ = (2/5) m (2a²)Wait, no, let me recall the formula for a prolate spheroid. A prolate spheroid is an ellipsoid with two equal semi-minor axes and one semi-major axis. The moments of inertia are:I₁ = I₂ = (1/5) m (4a² + b²)I₃ = (1/5) m (2b² + 2a²)Wait, no, maybe I should derive it.The moment of inertia for a prolate spheroid (ellipsoid with semi-axes a, a, b) about the major axis (b) is:I₃ = (2/5) m a²And about the minor axes (a):I₁ = I₂ = (1/5) m (2a² + b²)Wait, let me check that.The general formula for the moment of inertia of an ellipsoid with semi-axes a, b, c is:I₁ = (1/5) m (b² + c²)I₂ = (1/5) m (a² + c²)I₃ = (1/5) m (a² + b²)But in our case, it's a prolate spheroid, so a = b, and c is the semi-major axis. Wait, no, actually, in a prolate spheroid, the two equal axes are the semi-minor axes, and the third is the semi-major axis.So, if the semi-major axis is 14 cm (let's denote it as c = 14 cm), and the semi-minor axes are 7 cm (a = b = 7 cm).Therefore, the moments of inertia would be:I₁ = I₂ = (1/5) m (b² + c²) = (1/5) m (7² + 14²) = (1/5) m (49 + 196) = (1/5) m (245) = 49 mI₃ = (1/5) m (a² + b²) = (1/5) m (7² + 7²) = (1/5) m (49 + 49) = (1/5) m (98) = 19.6 mWait, but that seems off because I₁ and I₂ should be larger than I₃ since they are about the minor axes. Wait, no, actually, for a prolate spheroid, the moment of inertia about the major axis is smaller than about the minor axes because the mass is distributed further from the axis for the minor axes.Wait, let me think again. The moment of inertia is larger when the mass is further from the axis. So, for a prolate spheroid, the semi-major axis is longer, so the moment of inertia about the major axis should be larger than about the minor axes. Wait, no, actually, no. Because for a prolate spheroid, the major axis is longer, so when rotating about the major axis, the mass is distributed closer to the axis, so the moment of inertia is smaller. Whereas when rotating about the minor axes, the mass is distributed further out, so the moment of inertia is larger.Wait, that makes sense. So, I₁ and I₂ (about the minor axes) are larger than I₃ (about the major axis). So, in our case, I₁ = I₂ = (1/5) m (c² + b²) where c is the major axis? Wait, no, I think I got the axes mixed up.Wait, let's clarify. The prolate spheroid has two equal semi-minor axes (a) and one semi-major axis (b). So, the moments of inertia are:I₁ = I₂ = (1/5) m (b² + a²)I₃ = (1/5) m (2a²)Wait, no, let me refer to the standard formula.For an ellipsoid with semi-axes a, b, c, the moments of inertia are:I₁ = (1/5) m (b² + c²)I₂ = (1/5) m (a² + c²)I₃ = (1/5) m (a² + b²)But in a prolate spheroid, two axes are equal, say a = b, and c is the major axis. So, substituting a = b, we get:I₁ = I₂ = (1/5) m (a² + c²)I₃ = (1/5) m (2a²)So, in our case, the semi-major axis is 14 cm (c = 14 cm), and the semi-minor axes are 7 cm (a = 7 cm). Therefore:I₁ = I₂ = (1/5) m (7² + 14²) = (1/5) m (49 + 196) = (1/5) m (245) = 49 mI₃ = (1/5) m (2 * 7²) = (1/5) m (98) = 19.6 mWait, so I₁ and I₂ are 49 m, and I₃ is 19.6 m. That makes sense because I₁ and I₂ are about the minor axes, so they are larger.But in the problem, it says "the moments of inertia I₁ and I₃ along the semi-minor and semi-major axes, respectively". So, I₁ is along the semi-minor axis, which is 7 cm, and I₃ is along the semi-major axis, 14 cm.So, I₁ = 49 m, I₃ = 19.6 m.Therefore, going back to the rotational dynamics.Since the football is rotating about its principal axis (semi-major axis), ω₁ = ω₂ = 0, ω₃ = ω.Therefore, Euler's equations simplify to:I₁ * (dω₁/dt) = τ₁I₂ * (dω₂/dt) = τ₂I₃ * (dω₃/dt) = τ₃But since ω₁ and ω₂ are zero, and assuming no external torques in those directions (τ₁ = τ₂ = 0), then dω₁/dt = dω₂/dt = 0. So, the angular velocities in those directions remain zero.Therefore, the only equation we need is for ω₃:I₃ * (dω/dt) = τSo, the equation governing the rotational dynamics is:I₃ * (dω/dt) = τBut the problem mentions to account for the aerodynamic torque τ. So, perhaps τ is a function of ω or other variables. But since the problem doesn't specify the form of τ, I think we can just write the equation as:I₃ * (dω/dt) = τBut maybe we need to express τ in terms of other variables. For example, in aerodynamics, the torque can be related to the angular velocity and the drag force.Alternatively, perhaps the torque is due to the Magnus effect, which is proportional to the angular velocity and the velocity of the football. The Magnus torque is given by τ = k * ω × v, where k is a constant depending on the aerodynamics.But since the football is moving through the air, its velocity v is a vector, and the angular velocity ω is also a vector. The cross product would give the direction of the torque.But in this case, since the football is rotating about its principal axis (semi-major axis), and assuming it's moving along that axis, then v is along ω, so the cross product would be zero. Therefore, the Magnus torque would be zero.Wait, that can't be right because the Magnus effect causes a lateral force when the velocity and angular velocity are perpendicular. If they are parallel, there's no Magnus force or torque.So, in that case, maybe the aerodynamic torque is zero? Or perhaps it's due to other factors, like the pressure distribution causing a torque.Alternatively, maybe the torque is due to the difference in drag on the front and back of the football as it spins. If the football is spinning, the front and back might experience different drag forces, leading to a torque.But I'm not sure about the exact form. Since the problem just mentions to account for the aerodynamic torque τ, perhaps we can leave it as τ in the equation.Therefore, the equation governing the rotational dynamics is:I₃ * (dω/dt) = τBut let me check if there's any other torque. If the football is moving through the air, there might be a damping torque proportional to the angular velocity, similar to how linear drag is proportional to velocity.So, perhaps τ = -k * ω, where k is a damping coefficient. But the problem doesn't specify, so maybe we can just write the equation as:I₃ * (dω/dt) = τSo, that's part 1.Now, moving on to part 2: The student throws the ball with an initial velocity of 20 m/s at 30 degrees to the horizontal. We need to derive the equations of motion for the football's trajectory, considering air resistance modeled by F_d = -c_d v², where c_d = 0.1 kg/m, and g = 9.81 m/s².Okay, projectile motion with air resistance. Without air resistance, it's straightforward, but with quadratic drag, it's more complex.The equations of motion will involve both the x and y components of velocity, considering the drag force.First, let's set up the coordinate system. Let's assume the x-axis is horizontal, and the y-axis is vertical.The initial velocity is 20 m/s at 30 degrees, so:v₀x = 20 * cos(30°) = 20 * (√3/2) ≈ 17.32 m/sv₀y = 20 * sin(30°) = 20 * 0.5 = 10 m/sThe drag force is given by F_d = -c_d v², where v is the velocity vector. Since drag opposes the motion, it acts opposite to the velocity direction.Therefore, the drag force components are:F_x = -c_d v² * (v_x / v)F_y = -c_d v² * (v_y / v) - m gWait, no, the drag force is F_d = -c_d v² * hat{v}, where hat{v} is the unit vector in the direction of velocity. So, in components:F_x = -c_d v² * (v_x / v)F_y = -c_d v² * (v_y / v) - m gBut since F = m a, we can write the equations of motion as:m * dv_x/dt = -c_d v² * (v_x / v)m * dv_y/dt = -c_d v² * (v_y / v) - m gSimplifying, we can divide both sides by m:dv_x/dt = -(c_d / m) v² (v_x / v)dv_y/dt = -(c_d / m) v² (v_y / v) - gBut v² = v_x² + v_y², so v = sqrt(v_x² + v_y²)Therefore, the equations become:dv_x/dt = -(c_d / m) v_x vdv_y/dt = -(c_d / m) v_y v - gWhere v = sqrt(v_x² + v_y²)But the problem doesn't specify the mass of the football. Hmm, that's an issue. Maybe we can keep it in terms of m, or perhaps assume it's a standard football. But since the problem doesn't give the mass, maybe we can leave it as m.Alternatively, perhaps the drag force is given as F_d = -c_d v², so the acceleration is a = F/m = - (c_d / m) v² * hat{v}So, the equations are:dv_x/dt = -(c_d / m) v_x vdv_y/dt = -(c_d / m) v_y v - gYes, that seems right.But let me check the units. c_d is given as 0.1 kg/m. So, c_d has units of kg/m.v has units of m/s, so v² is m²/s².Therefore, F_d = c_d v² has units kg/m * m²/s² = kg m/s² = N, which is correct.So, the equations are correct.Therefore, the equations of motion are:dv_x/dt = -(c_d / m) v_x vdv_y/dt = -(c_d / m) v_y v - gWith initial conditions:v_x(0) = 20 cos(30°) = 10√3 ≈ 17.32 m/sv_y(0) = 20 sin(30°) = 10 m/sAnd the position equations would be:dx/dt = v_xdy/dt = v_yWith initial positions x(0) = 0, y(0) = 0 (assuming it's thrown from the origin).But the problem only asks for the equations of motion, not to solve them, so we can stop here.Wait, but maybe we can write them in a more compact form. Let me see.Alternatively, since v = sqrt(v_x² + v_y²), we can write:dv_x/dt = -(c_d / m) v_x sqrt(v_x² + v_y²)dv_y/dt = -(c_d / m) v_y sqrt(v_x² + v_y²) - gYes, that's the complete set of differential equations governing the trajectory.But perhaps we can write them in terms of dimensionless variables or make substitutions, but I think that's beyond the scope here.So, to summarize part 2, the equations of motion are:dv_x/dt = -(c_d / m) v_x sqrt(v_x² + v_y²)dv_y/dt = -(c_d / m) v_y sqrt(v_x² + v_y²) - gWith c_d = 0.1 kg/m, g = 9.81 m/s², and initial velocities as above.I think that's it. Let me just recap.Part 1: The rotational dynamics equation is I₃ * (dω/dt) = τ, where I₃ is the moment of inertia about the semi-major axis, and τ is the aerodynamic torque.Part 2: The equations of motion considering air resistance are the two differential equations above for v_x and v_y.I think that's all.</think>

🧠:<think>Okay, so I have this problem about David and Tom betting on their college football teams, the Florida Gators and the Miami Hurricanes, respectively. They're planning to watch a 5-game series, and they want to model the probabilities of their teams winning. There are two parts to this problem, and I need to figure both out step by step.Starting with part 1: Historically, the Florida Gators have a 60% chance of winning any individual game against the Miami Hurricanes. Let ( X ) be the random variable representing the number of games the Florida Gators win out of the 5-game series. I need to find the probability mass function (PMF) of ( X ) and determine the probability that the Florida Gators win exactly 3 games in the series.Hmm, okay. So, first, I remember that when dealing with the number of successes in a series of independent trials, we can model this using the binomial distribution. Each game is an independent trial with two possible outcomes: win or loss. Since the probability of winning each game is constant at 60%, this fits the binomial model.The PMF of a binomial distribution is given by the formula:[P(X = k) = C(n, k) times p^k times (1 - p)^{n - k}]Where:- ( n ) is the number of trials (in this case, 5 games),- ( k ) is the number of successes (Gators wins),- ( p ) is the probability of success on a single trial (60% or 0.6),- ( C(n, k) ) is the combination of ( n ) things taken ( k ) at a time.So, for part 1, the PMF of ( X ) is just this formula with ( n = 5 ) and ( p = 0.6 ). Therefore, the PMF is:[P(X = k) = C(5, k) times (0.6)^k times (0.4)^{5 - k}]for ( k = 0, 1, 2, 3, 4, 5 ).Now, they specifically ask for the probability that the Florida Gators win exactly 3 games. So, I need to calculate ( P(X = 3) ).Plugging into the formula:First, calculate the combination ( C(5, 3) ). That's the number of ways to choose 3 successes out of 5 trials.[C(5, 3) = frac{5!}{3!(5 - 3)!} = frac{5 times 4 times 3!}{3! times 2!} = frac{5 times 4}{2 times 1} = 10]So, ( C(5, 3) = 10 ).Next, ( (0.6)^3 = 0.6 times 0.6 times 0.6 = 0.216 ).Then, ( (0.4)^{5 - 3} = (0.4)^2 = 0.16 ).Multiplying all together:[P(X = 3) = 10 times 0.216 times 0.16]Let me compute that step by step.First, 10 multiplied by 0.216 is 2.16.Then, 2.16 multiplied by 0.16. Let's see:0.16 times 2 is 0.32, and 0.16 times 0.16 is 0.0256. Wait, no, that's not the right way. Wait, 2.16 times 0.16.Alternatively, 2.16 * 0.16:2 * 0.16 = 0.320.16 * 0.16 = 0.0256Wait, no, that's not correct. Wait, 2.16 is 2 + 0.16. So, 2 * 0.16 is 0.32, and 0.16 * 0.16 is 0.0256. So, adding together, 0.32 + 0.0256 = 0.3456.Wait, but that seems off because 2.16 * 0.16 is actually:2.16x0.16------Multiply 2.16 by 0.1: 0.216Multiply 2.16 by 0.06: 0.1296Add them together: 0.216 + 0.1296 = 0.3456Yes, that's correct. So, 2.16 * 0.16 = 0.3456.Therefore, ( P(X = 3) = 0.3456 ).So, the probability that the Florida Gators win exactly 3 games is 0.3456, or 34.56%.Wait, let me just double-check my calculations to make sure I didn't make a mistake.First, ( C(5, 3) = 10 ), that's correct.( (0.6)^3 = 0.216 ), correct.( (0.4)^2 = 0.16 ), correct.10 * 0.216 = 2.16, correct.2.16 * 0.16: Let's do it another way. 2.16 * 0.16 is equal to (2 + 0.16) * 0.16 = 2*0.16 + 0.16*0.16 = 0.32 + 0.0256 = 0.3456. Yep, that's correct.So, I think that's solid.Moving on to part 2: David and Tom decide to add a twist to their bet. They agree that if their respective teams win at least 3 games in the series, they will each donate 100 to a charity. So, if the Florida Gators win at least 3 games, David donates 100, and if the Miami Hurricanes win at least 3 games, Tom donates 100. The question is to calculate the expected total donation amount to charity.Given that ( P(X geq 3) ) is the probability that the Florida Gators win at least 3 games, and ( P(Y geq 3) ) is the probability that the Miami Hurricanes win at least 3 games, where ( Y ) is the random variable representing the number of games the Miami Hurricanes win, with a 40% win probability for each game.So, the expected total donation is the sum of the expected donations from David and Tom. Since each donates 100 if their team wins at least 3 games, the expected donation is:[E[text{Total Donation}] = 100 times P(X geq 3) + 100 times P(Y geq 3)]So, I need to compute ( P(X geq 3) ) and ( P(Y geq 3) ), then multiply each by 100 and sum them up.First, let's compute ( P(X geq 3) ). Since ( X ) is the number of Gators wins, which follows a binomial distribution with ( n = 5 ) and ( p = 0.6 ).Similarly, ( Y ) is the number of Hurricanes wins, which is also binomial with ( n = 5 ) and ( p = 0.4 ). Wait, but is that correct? Because in a 5-game series, the number of wins for the Hurricanes would be ( Y = 5 - X ), right? So, if the Gators win ( X ) games, the Hurricanes win ( Y = 5 - X ) games.Therefore, ( Y ) is not an independent binomial variable with ( p = 0.4 ); rather, it's directly dependent on ( X ). However, in this case, since each game is independent, and the probability of the Hurricanes winning a game is 40%, ( Y ) does indeed follow a binomial distribution with ( n = 5 ) and ( p = 0.4 ). So, I can model ( Y ) as a separate binomial variable.Therefore, ( P(Y geq 3) ) is the probability that the Hurricanes win at least 3 games, which is the same as ( P(X leq 2) ) because ( Y = 5 - X ). So, ( P(Y geq 3) = P(X leq 2) ).But let's proceed step by step.First, compute ( P(X geq 3) ). This is the probability that the Gators win 3, 4, or 5 games.Similarly, ( P(Y geq 3) ) is the probability that the Hurricanes win 3, 4, or 5 games.But since each game is either a Gator win or a Hurricane win, and the total number of games is 5, the events ( X geq 3 ) and ( Y geq 3 ) are mutually exclusive? Wait, no, actually, they can overlap. Wait, no, because if ( X geq 3 ), then ( Y = 5 - X leq 2 ), so they are mutually exclusive. Wait, is that correct?Wait, if ( X geq 3 ), then ( Y = 5 - X leq 2 ). Similarly, if ( Y geq 3 ), then ( X leq 2 ). So, the events ( X geq 3 ) and ( Y geq 3 ) are mutually exclusive and collectively exhaustive? No, because the total probability space is covered by ( X + Y = 5 ). So, the events ( X geq 3 ) and ( Y geq 3 ) cannot happen at the same time, but they don't cover the entire probability space because there's also the possibility that both ( X ) and ( Y ) are less than 3, which would mean the series ends with a tie in wins, but since it's 5 games, which is odd, one team must have at least 3 wins. Wait, no, 5 is odd, so one team must have 3 or more wins, and the other has 2 or fewer. So, actually, the events ( X geq 3 ) and ( Y geq 3 ) are mutually exclusive and collectively exhaustive. Because in a 5-game series, one team must have at least 3 wins, so either ( X geq 3 ) or ( Y geq 3 ), but not both.Therefore, ( P(X geq 3) + P(Y geq 3) = 1 ). So, if I compute ( P(X geq 3) ), then ( P(Y geq 3) = 1 - P(X geq 3) ).But let me verify that.Wait, in a 5-game series, the number of wins for each team must add up to 5. So, if the Gators have 3 wins, the Hurricanes have 2. If the Gators have 4 wins, the Hurricanes have 1. If the Gators have 5 wins, the Hurricanes have 0. Similarly, if the Hurricanes have 3 wins, the Gators have 2, and so on.Therefore, the events ( X geq 3 ) and ( Y geq 3 ) are indeed mutually exclusive because they can't both have 3 or more wins in a 5-game series. Also, since one of them must have at least 3 wins, these two events cover the entire probability space. Therefore, ( P(X geq 3) + P(Y geq 3) = 1 ).So, if I compute ( P(X geq 3) ), then ( P(Y geq 3) = 1 - P(X geq 3) ).But let's compute both just to make sure.First, compute ( P(X geq 3) ). That is:[P(X geq 3) = P(X = 3) + P(X = 4) + P(X = 5)]We already computed ( P(X = 3) = 0.3456 ).Now, let's compute ( P(X = 4) ) and ( P(X = 5) ).Starting with ( P(X = 4) ):Using the binomial formula:[P(X = 4) = C(5, 4) times (0.6)^4 times (0.4)^{1}]Compute each part:( C(5, 4) = 5 ).( (0.6)^4 = 0.6 times 0.6 times 0.6 times 0.6 ).Let me compute that:0.6^2 = 0.360.36^2 = 0.1296So, 0.6^4 = 0.1296.Then, ( (0.4)^1 = 0.4 ).So, multiplying together:5 * 0.1296 * 0.4First, 5 * 0.1296 = 0.648Then, 0.648 * 0.4 = 0.2592So, ( P(X = 4) = 0.2592 ).Next, ( P(X = 5) ):[P(X = 5) = C(5, 5) times (0.6)^5 times (0.4)^0]Compute each part:( C(5, 5) = 1 ).( (0.6)^5 = 0.6 times 0.6 times 0.6 times 0.6 times 0.6 ).We know 0.6^4 is 0.1296, so 0.1296 * 0.6 = 0.07776.( (0.4)^0 = 1 ).So, multiplying together:1 * 0.07776 * 1 = 0.07776.Therefore, ( P(X = 5) = 0.07776 ).Now, summing up ( P(X = 3) + P(X = 4) + P(X = 5) ):0.3456 + 0.2592 + 0.07776Let me compute that step by step.0.3456 + 0.2592 = 0.60480.6048 + 0.07776 = 0.68256So, ( P(X geq 3) = 0.68256 ).Therefore, ( P(Y geq 3) = 1 - 0.68256 = 0.31744 ).Alternatively, I can compute ( P(Y geq 3) ) directly by considering the binomial distribution for ( Y ) with ( p = 0.4 ).Let me do that to verify.Compute ( P(Y geq 3) = P(Y = 3) + P(Y = 4) + P(Y = 5) ).First, ( P(Y = 3) ):[P(Y = 3) = C(5, 3) times (0.4)^3 times (0.6)^2]Compute each part:( C(5, 3) = 10 ).( (0.4)^3 = 0.064 ).( (0.6)^2 = 0.36 ).Multiplying together:10 * 0.064 * 0.36First, 10 * 0.064 = 0.640.64 * 0.36 = 0.2304So, ( P(Y = 3) = 0.2304 ).Next, ( P(Y = 4) ):[P(Y = 4) = C(5, 4) times (0.4)^4 times (0.6)^1]Compute each part:( C(5, 4) = 5 ).( (0.4)^4 = 0.0256 ).( (0.6)^1 = 0.6 ).Multiplying together:5 * 0.0256 * 0.6First, 5 * 0.0256 = 0.1280.128 * 0.6 = 0.0768So, ( P(Y = 4) = 0.0768 ).Next, ( P(Y = 5) ):[P(Y = 5) = C(5, 5) times (0.4)^5 times (0.6)^0]Compute each part:( C(5, 5) = 1 ).( (0.4)^5 = 0.01024 ).( (0.6)^0 = 1 ).Multiplying together:1 * 0.01024 * 1 = 0.01024.So, ( P(Y = 5) = 0.01024 ).Now, summing up ( P(Y = 3) + P(Y = 4) + P(Y = 5) ):0.2304 + 0.0768 + 0.01024Let's compute:0.2304 + 0.0768 = 0.30720.3072 + 0.01024 = 0.31744So, ( P(Y geq 3) = 0.31744 ), which matches our earlier calculation of ( 1 - P(X geq 3) = 1 - 0.68256 = 0.31744 ). So, that checks out.Therefore, the expected total donation is:[E[text{Total Donation}] = 100 times P(X geq 3) + 100 times P(Y geq 3) = 100 times 0.68256 + 100 times 0.31744]Simplify:[= 100 times (0.68256 + 0.31744) = 100 times 1 = 100]Wait, that's interesting. So, the expected total donation is 100.But let me think about that. Since in every possible outcome, either the Gators win at least 3 games or the Hurricanes win at least 3 games, but not both, the total donation will always be 100, because exactly one of them will donate 100. Therefore, the expected donation is 100.But let me verify that.Wait, actually, no. Because in reality, if both teams could somehow win exactly 2.5 games, but since the number of games is 5, which is odd, one team must have at least 3 wins. Therefore, in every possible outcome, exactly one of the two teams will have at least 3 wins, so exactly one of David or Tom will donate 100. Therefore, the total donation is always 100, regardless of the outcome. Therefore, the expected total donation is 100.But wait, that seems a bit counterintuitive because the probabilities are different. The Gators have a higher chance of winning each game, so David is more likely to donate, but Tom is less likely. However, regardless of who donates, the amount donated is always 100. Therefore, the expected total donation is 100.Wait, but let me think again. If the donation is 100 from each if their respective teams win at least 3 games, but in reality, only one of them donates each time. So, the total donation is either 100 (if Gators win at least 3) or 100 (if Hurricanes win at least 3). Wait, no, actually, no. Wait, no, if Gators win at least 3, David donates 100, and Tom doesn't. If Hurricanes win at least 3, Tom donates 100, and David doesn't. So, in either case, only one person donates 100, so the total donation is always 100. Therefore, the expected total donation is 100, because regardless of the outcome, the total donated is 100.Therefore, the expected total donation is 100.But let me compute it the long way to make sure.Compute ( E[text{Total Donation}] = 100 times P(X geq 3) + 100 times P(Y geq 3) ).We have ( P(X geq 3) = 0.68256 ) and ( P(Y geq 3) = 0.31744 ).So, 100 * 0.68256 = 68.256100 * 0.31744 = 31.744Adding them together: 68.256 + 31.744 = 100.Yes, that's correct. So, the expected total donation is 100.Therefore, regardless of the probabilities, since the total donation is always 100, the expectation is 100.But wait, that seems a bit strange because if the probabilities were different, say, if both teams had a 50% chance, then the expected donation would still be 100, because one of them would always donate. So, in this case, even though the Gators have a higher chance, the expectation is still 100 because it's a guaranteed 100 donation regardless of who wins.Therefore, the answer is 100.But let me think again. Suppose in a different scenario, if both teams could somehow both win at least 3 games, then the total donation could be 200. But in a 5-game series, that's impossible because one team must have at least 3 wins, and the other has at most 2. Therefore, the total donation is always 100, so the expectation is 100.Therefore, the expected total donation amount to charity is 100.So, summarizing:1. The PMF of ( X ) is binomial with ( n = 5 ) and ( p = 0.6 ), and ( P(X = 3) = 0.3456 ).2. The expected total donation is 100.Final Answer1. The probability that the Florida Gators win exactly 3 games is boxed{0.3456}.2. The expected total donation amount to charity is boxed{100} dollars.</think>

🧠:<think>Okay, so I have this problem here about a Broadway enthusiast who has 15 original cast recordings, each with a distinct prime number of tracks. They also have a Michael Jackson album with 25 tracks. There are two parts to the problem.Starting with part 1: I need to determine the minimum possible number of tracks across all 15 Broadway recordings combined. Each recording has a different prime number of tracks, so I need to find 15 distinct prime numbers and sum them up, aiming for the smallest total possible.Hmm, primes. The smallest prime is 2, then 3, 5, 7, 11, and so on. So, to get the minimum total, I should use the first 15 prime numbers. Let me list them out:1. 22. 33. 54. 75. 116. 137. 178. 199. 2310. 2911. 3112. 3713. 4114. 4315. 47Wait, is that right? Let me double-check. The primes start at 2, then 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Yeah, that's 15 primes. So, now I need to sum these up.Let me add them step by step:Start with 2 + 3 = 55 + 5 = 1010 + 7 = 1717 + 11 = 2828 + 13 = 4141 + 17 = 5858 + 19 = 7777 + 23 = 100100 + 29 = 129129 + 31 = 160160 + 37 = 197197 + 41 = 238238 + 43 = 281281 + 47 = 328So, the total is 328 tracks. Is that correct? Let me verify the addition:2 + 3 = 55 + 5 = 1010 + 7 = 1717 + 11 = 2828 + 13 = 4141 + 17 = 5858 + 19 = 7777 + 23 = 100100 + 29 = 129129 + 31 = 160160 + 37 = 197197 + 41 = 238238 + 43 = 281281 + 47 = 328Yes, that seems to add up correctly. So, the minimum possible number of tracks is 328.Moving on to part 2: The enthusiast creates a playlist by selecting exactly one track from each Broadway recording and one track from the Michael Jackson album. The probability that the selected track from any given recording is the "highlight" is 1/5. We need to find the probability that the playlist contains exactly 5 "highlight" tracks from the Broadway recordings and the "highlight" track from the Michael Jackson album.Alright, so this is a probability problem involving combinations and independent events.First, let's break it down. There are 15 Broadway recordings, each with a 1/5 chance of selecting the highlight track. The Michael Jackson album has a 1/25 chance of selecting its highlight track, since there are 25 tracks.We need exactly 5 highlights from the Broadway shows and 1 highlight from Michael Jackson.So, the probability is the product of two probabilities:1. The probability of selecting exactly 5 highlights from the 15 Broadway recordings.2. The probability of selecting the highlight from the Michael Jackson album.Let me handle each part separately.First, for the Broadway recordings: This is a binomial probability scenario. The number of ways to choose 5 highlights out of 15, multiplied by the probability of getting 5 highlights and 10 non-highlights.The formula for binomial probability is:P(k) = C(n, k) * p^k * (1-p)^(n-k)Where:- n = 15- k = 5- p = 1/5So, P(5) = C(15, 5) * (1/5)^5 * (4/5)^10I can compute this, but maybe I should leave it in terms of combinations for now.Next, the Michael Jackson album: There's a 1/25 chance of selecting the highlight track.So, the total probability is P(5) * (1/25)Therefore, the total probability is:C(15, 5) * (1/5)^5 * (4/5)^10 * (1/25)Simplify this expression.First, compute C(15, 5). That's 3003.So, 3003 * (1/5)^5 * (4/5)^10 * (1/25)Compute each part:(1/5)^5 = 1 / 3125(4/5)^10 is approximately, but let me compute it exactly.(4/5)^10 = (4^10)/(5^10) = 1048576 / 9765625So, putting it all together:3003 * (1 / 3125) * (1048576 / 9765625) * (1 / 25)Multiply all the constants:First, 3003 * 1048576 = ?Wait, that's a big number. Maybe we can compute it step by step.But perhaps it's better to write it as:3003 * (1 / 3125) * (1048576 / 9765625) * (1 / 25) = 3003 * 1048576 / (3125 * 9765625 * 25)Compute the denominator:3125 * 25 = 7812578125 * 9765625 = ?Wait, 78125 * 9765625. Let's compute that.But 78125 is 5^7, since 5^7 = 78125.9765625 is 5^10, since 5^10 = 9765625.So, 5^7 * 5^10 = 5^17Similarly, 3003 * 1048576 is 3003 * 2^20, since 1048576 = 2^20.So, numerator is 3003 * 2^20, denominator is 5^17.So, the probability is (3003 * 2^20) / (5^17)Alternatively, we can write it as (3003 / 5^17) * 2^20But maybe we can compute this as a decimal.Alternatively, perhaps we can compute it step by step.But let me see if I can compute 3003 * 1048576 first.3003 * 1,048,576Let me compute 3000 * 1,048,576 = 3,145,728,000Then, 3 * 1,048,576 = 3,145,728So, total is 3,145,728,000 + 3,145,728 = 3,148,873,728So, numerator is 3,148,873,728Denominator is 3125 * 9765625 * 25Compute 3125 * 25 = 78,12578,125 * 9,765,625Compute 78,125 * 9,765,625Let me note that 78,125 * 9,765,625 = 78,125 * (9,765,625)But 78,125 is 5^7, and 9,765,625 is 5^10, so 5^17, which is 762,939,453,125Wait, 5^1 = 55^2 =255^3=1255^4=6255^5=31255^6=15,6255^7=78,1255^8=390,6255^9=1,953,1255^10=9,765,6255^11=48,828,1255^12=244,140,6255^13=1,220,703,1255^14=6,103,515,6255^15=30,517,578,1255^16=152,587,890,6255^17=762,939,453,125Yes, so denominator is 762,939,453,125So, the probability is 3,148,873,728 / 762,939,453,125Let me compute this division.First, note that 3,148,873,728 ÷ 762,939,453,125This is approximately equal to:Let me compute 3,148,873,728 / 762,939,453,125 ≈ 0.004126Wait, let me compute it more accurately.Compute 3,148,873,728 ÷ 762,939,453,125Let me write both numbers in scientific notation.3,148,873,728 ≈ 3.148873728 x 10^9762,939,453,125 ≈ 7.62939453125 x 10^11So, dividing them:(3.148873728 x 10^9) / (7.62939453125 x 10^11) = (3.148873728 / 762.939453125) x 10^(-2)Compute 3.148873728 / 762.939453125 ≈ 0.004126So, 0.004126 x 10^(-2) = 0.00004126Wait, that can't be right because 3.148873728 / 762.939453125 is approximately 0.004126, and then times 10^(-2) would be 0.00004126.But let me check:Wait, 3.148873728 / 762.939453125 = approximately 0.004126But 3.148873728 / 762.939453125 is the same as 3.148873728 divided by 762.939453125.Compute 3.148873728 ÷ 762.939453125Well, 762.939453125 goes into 3.148873728 approximately 0.004126 times.Because 762.939453125 * 0.004126 ≈ 3.148873728Yes, that seems correct.So, the probability is approximately 0.00004126, which is 4.126 x 10^(-5)But let me express it as a fraction.We have 3,148,873,728 / 762,939,453,125Simplify numerator and denominator by dividing numerator and denominator by GCD(3,148,873,728, 762,939,453,125)But that might be complicated. Alternatively, note that 3,148,873,728 = 3003 * 1,048,576And 762,939,453,125 = 5^17So, the fraction is (3003 * 2^20) / 5^17Which is the exact form.Alternatively, we can write it as (3003 / 5^17) * 2^20But perhaps we can compute it as a decimal.Alternatively, let me compute 3,148,873,728 ÷ 762,939,453,125Let me do this division step by step.First, 762,939,453,125 goes into 3,148,873,728 how many times?Compute 762,939,453,125 x 0.004 = 3,051,757,812.5Subtract that from 3,148,873,728:3,148,873,728 - 3,051,757,812.5 = 97,115,915.5Now, 762,939,453,125 x 0.000126 ≈ 97,115,915.5Because 762,939,453,125 x 0.0001 = 76,293,945.3125So, 0.000126 would be approximately 76,293,945.3125 * 1.26 ≈ 96,180,000Which is close to 97,115,915.5So, total is approximately 0.004 + 0.000126 ≈ 0.004126So, the probability is approximately 0.004126, which is 0.4126%But wait, that seems low, but considering the number of combinations, maybe it's correct.Alternatively, let me think about it differently.The probability of getting exactly 5 highlights from Broadway is C(15,5)*(1/5)^5*(4/5)^10, which is approximately:C(15,5) is 3003(1/5)^5 = 1/3125 ≈ 0.00032(4/5)^10 ≈ 0.107374So, 3003 * 0.00032 * 0.107374 ≈ 3003 * 0.00003436 ≈ 0.1032So, approximately 10.32% chance for exactly 5 highlights from Broadway.Then, the probability of selecting the highlight from Michael Jackson is 1/25 = 0.04So, total probability is 0.1032 * 0.04 ≈ 0.004128, which is about 0.4128%Which matches our earlier calculation.So, approximately 0.4128% chance.But the question might want an exact fraction or a simplified form.Given that, the exact probability is (3003 * 2^20) / 5^17But let me compute 2^20 = 1,048,576And 5^17 = 762,939,453,125So, 3003 * 1,048,576 = 3,148,873,728So, the exact probability is 3,148,873,728 / 762,939,453,125We can simplify this fraction by dividing numerator and denominator by GCD(3,148,873,728, 762,939,453,125)But 3,148,873,728 is 3003 * 1,048,576And 762,939,453,125 is 5^17Since 3003 and 5^17 are co-prime (3003 is 3*7*11*13, none of which are factors of 5), and 1,048,576 is 2^20, which is also co-prime with 5^17.Therefore, the fraction cannot be simplified further.So, the exact probability is 3,148,873,728 / 762,939,453,125Alternatively, we can write it as 3003 * 2^20 / 5^17But perhaps the question expects the answer in terms of combinations and exponents, or as a decimal.Given that, I think either form is acceptable, but since it's a probability, it's often expressed as a decimal or a fraction.But since the numbers are large, maybe expressing it in terms of exponents is better.Alternatively, we can write it as:C(15,5) * (1/5)^5 * (4/5)^10 * (1/25) = 3003 * (1/5)^5 * (4/5)^10 * (1/25)Which is the exact form.Alternatively, we can write it as:3003 * 4^10 / (5^15 * 25) = 3003 * 4^10 / 5^17But 4^10 is 2^20, so it's the same as above.So, I think the exact probability is 3003 * 2^20 / 5^17, which is approximately 0.004126 or 0.4126%But to express it as a fraction, it's 3,148,873,728 / 762,939,453,125Simplify numerator and denominator by dividing both by 3,148,873,728 and 762,939,453,125 by their GCD.But since they are co-prime, as we saw earlier, the fraction is already in its simplest form.So, the probability is 3,148,873,728 / 762,939,453,125Alternatively, as a decimal, approximately 0.004126But let me check if I can write it as a reduced fraction.Wait, 3,148,873,728 divided by 762,939,453,125Let me see if both are divisible by 3.Sum of digits of numerator: 3+1+4+8+8+7+3+7+2+8 = 3+1=4, 4+4=8, 8+8=16, 16+7=23, 23+3=26, 26+7=33, 33+2=35, 35+8=43. 43 is not divisible by 3.Denominator: 7+6+2+9+3+9+4+5+3+1+2+5 = 7+6=13, 13+2=15, 15+9=24, 24+3=27, 27+9=36, 36+4=40, 40+5=45, 45+3=48, 48+1=49, 49+2=51, 51+5=56. 56 is not divisible by 3.So, not divisible by 3.Check divisibility by 2: numerator is even, denominator is odd, so no.Divisibility by 5: numerator ends with 8, so no. Denominator ends with 5, so yes, but numerator doesn't, so no.So, the fraction cannot be reduced further.Therefore, the exact probability is 3,148,873,728 / 762,939,453,125Alternatively, we can write it as approximately 0.004126 or 0.4126%But I think the question might prefer the exact form, so I'll go with the fraction.So, summarizing:1. The minimum total number of tracks is 328.2. The probability is 3,148,873,728 / 762,939,453,125, which is approximately 0.004126 or 0.4126%But let me check if I made any mistakes in the calculations.Wait, in the binomial probability, I used C(15,5)*(1/5)^5*(4/5)^10, which is correct.Then multiplied by 1/25 for the Michael Jackson album.Yes, that seems right.So, the calculations seem correct.</think>

🧠:<think>Alright, so I have two probability problems to solve here. Let me take them one at a time.Starting with the first one: It's about a person who watches "Poker Face" and predicts plot twists. They predict 10 plot twists per episode with an accuracy rate that follows a binomial distribution, where the probability of a correct prediction is p = 0.75. I need to find the probability that they correctly predict at least 8 out of 10 plot twists in a single episode.Hmm, okay. So, binomial distribution. The formula for the probability of getting exactly k successes in n trials is given by:P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)Where C(n, k) is the combination of n things taken k at a time. Since the question is about at least 8 correct predictions, that means I need to calculate the probabilities for 8, 9, and 10 correct predictions and then sum them up.So, let me write that down:P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)Calculating each term:First, P(X = 8):C(10, 8) = 45p^8 = (0.75)^8(1 - p)^(10 - 8) = (0.25)^2So, P(X = 8) = 45 * (0.75)^8 * (0.25)^2Similarly, P(X = 9):C(10, 9) = 10p^9 = (0.75)^9(1 - p)^(10 - 9) = (0.25)^1So, P(X = 9) = 10 * (0.75)^9 * (0.25)^1And P(X = 10):C(10, 10) = 1p^10 = (0.75)^10(1 - p)^(10 - 10) = (0.25)^0 = 1So, P(X = 10) = 1 * (0.75)^10 * 1 = (0.75)^10Now, I need to compute each of these. Let me calculate each term step by step.First, let's compute (0.75)^8:0.75^2 = 0.56250.75^4 = (0.5625)^2 ≈ 0.316406250.75^8 = (0.31640625)^2 ≈ 0.0999755859Similarly, (0.75)^9 = (0.75)^8 * 0.75 ≈ 0.0999755859 * 0.75 ≈ 0.0749816894And (0.75)^10 = (0.75)^9 * 0.75 ≈ 0.0749816894 * 0.75 ≈ 0.056236267Now, (0.25)^2 = 0.0625(0.25)^1 = 0.25So, plugging back into each probability:P(X = 8) = 45 * 0.0999755859 * 0.0625Let me compute 45 * 0.0999755859 first. 45 * 0.1 is 4.5, so 45 * 0.0999755859 ≈ 45 * 0.1 - 45 * 0.000024414 ≈ 4.5 - 0.0011 ≈ 4.4989Then, 4.4989 * 0.0625 ≈ 0.28118125So, P(X = 8) ≈ 0.28118125Next, P(X = 9) = 10 * 0.0749816894 * 0.2510 * 0.0749816894 = 0.7498168940.749816894 * 0.25 = 0.1874542235So, P(X = 9) ≈ 0.1874542235Lastly, P(X = 10) = 0.056236267So, adding them up:0.28118125 + 0.1874542235 + 0.056236267 ≈0.28118125 + 0.1874542235 = 0.46863547350.4686354735 + 0.056236267 ≈ 0.5248717405So, approximately 0.5249, or 52.49% chance.Wait, that seems a bit high, but considering the high probability of success (0.75), it's plausible.Alternatively, maybe I can use the binomial formula in another way or check with a calculator.Alternatively, maybe using the complement, but since it's only 3 terms, it's manageable.Alternatively, perhaps using the normal approximation? But since n is 10, which is small, the normal approximation might not be accurate. So, better stick with the exact binomial calculation.So, I think my calculation is correct.Now, moving on to the second problem.Assume that the number of hours they spend each day, T, follows a normal distribution with mean μ = 2 hours and standard deviation σ = 0.5 hours.We need to find the probability that over a 30-day period, they spend more than 65 hours in total on watching and predicting.So, total hours over 30 days is the sum of 30 independent normal variables, each with mean 2 and standard deviation 0.5.The sum of normal variables is also normal, with mean μ_total = n * μ = 30 * 2 = 60 hours.The variance of the sum is n * σ^2 = 30 * (0.5)^2 = 30 * 0.25 = 7.5Therefore, standard deviation of the total is sqrt(7.5) ≈ 2.7386 hours.So, we have the total hours, let's denote it as S, follows N(60, 7.5).We need P(S > 65).To find this probability, we can standardize S.Z = (S - μ_total) / σ_total = (65 - 60) / sqrt(7.5) ≈ 5 / 2.7386 ≈ 1.826So, Z ≈ 1.826We need P(Z > 1.826). Since standard normal tables give P(Z < z), we can find P(Z < 1.826) and subtract from 1.Looking up 1.826 in the standard normal table. Let me recall that 1.82 corresponds to about 0.9656, and 1.83 corresponds to about 0.9664.Since 1.826 is between 1.82 and 1.83, let's interpolate.Difference between 1.82 and 1.83 is 0.01 in z, which corresponds to a difference of 0.9664 - 0.9656 = 0.0008 in probability.1.826 is 0.006 above 1.82, so proportionally, 0.006 / 0.01 = 0.6 of the difference.So, 0.9656 + 0.6 * 0.0008 = 0.9656 + 0.00048 = 0.96608Therefore, P(Z < 1.826) ≈ 0.9661Thus, P(Z > 1.826) = 1 - 0.9661 = 0.0339, or approximately 3.39%.Alternatively, using a calculator, the exact value can be found, but since 1.826 is approximately 1.83, which is 0.9664, so 1 - 0.9664 = 0.0336, which is about 3.36%.So, approximately 3.36% to 3.39% chance.Therefore, the probability is roughly 3.37%.Wait, but let me verify my calculations.Total mean is 60, total standard deviation is sqrt(30*(0.5)^2) = sqrt(7.5) ≈ 2.7386.So, (65 - 60)/2.7386 ≈ 5 / 2.7386 ≈ 1.826.Yes, that's correct.Looking up 1.826 in standard normal table:Using a more precise method, perhaps using linear approximation between 1.82 and 1.83.At z = 1.82, cumulative probability is 0.9656At z = 1.83, cumulative probability is 0.9664So, the difference is 0.0008 over 0.01 z.So, for 0.006 beyond 1.82, the cumulative probability increases by 0.006 / 0.01 * 0.0008 = 0.00048So, total cumulative probability is 0.9656 + 0.00048 = 0.96608Thus, P(Z > 1.826) = 1 - 0.96608 = 0.03392, approximately 3.39%.Alternatively, using a calculator, perhaps more accurately:Using the standard normal distribution, the exact value can be found using the error function or a calculator.Alternatively, using a z-table with more decimal places.But since I don't have a calculator here, I think 3.39% is a reasonable approximation.Alternatively, if I use the precise z-score of 1.826, perhaps the exact value is around 0.0339.So, approximately 3.39% chance.Therefore, summarizing:1. Probability of at least 8 correct predictions in an episode is approximately 52.49%.2. Probability of spending more than 65 hours over 30 days is approximately 3.39%.I think that's it.Final Answer1. The probability is boxed{0.5249}.2. The probability is boxed{0.0339}.</think>